2011 Multi-University Training Contest 7 - Host by ECNU
AC: F I. rank 40/88.
开场看了F发现是个简单的DP，随便写了一下WA,,,发现把样例倒着输就过不了了。。。原来是忘了最后的时候开始上课的话可能上不了多久。。。
想到一个简洁的状态方程，然后以为是单调队列优化，突然发现好像只需要求个最小值就行了。。。改改AC了。。
看榜发现I过了很多，看了一下就是个同色三角形裸题。。A了然后就没有然后了。。。
看了一下B。第一感觉KMP，感觉通配符的情况不大好处理。。。没学AC自动机。。。
C。求多边形的对称轴。。。但是枚举对称轴好像会TLE？不敢写。。。赛后发现数据水暴力也能过？另外标称是后缀数组。。膜。。。
D。感觉好像是化一化公式应该能看出什么性质？
H。树形DP？ J。dance link？
B.Wildcard(贪心+hash)
这题很骚。。就是bzoj 3507的原题。然后唯一不同的是，这题是判断通配符串是否为目标串的子串。。。
把通配符串两端各加一个*就变成了bzoj 3507...
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-4 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; inline int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=100010; //Code begin... char str[N], ss[N]; int hs[N], shs[N], xing[55], pos, bg, ed; struct Node{int x, len; bool flag;}node[55]; int get_hash(int l, int r){return shs[r]-shs[l-1]*hs[r-l+1];} bool check(int be, int L){ int sum=0; FOR(i,1,L) if (node[i].flag) sum+=node[i].x; else sum+=node[i].len; if (be+sum>ed+1) return false; FOR(i,1,L) { if (node[i].flag) be+=node[i].x; else { if (get_hash(be,be+node[i].len-1)!=node[i].x) return false; be+=node[i].len; } } return true; } bool sol(){ int num, sum, l, cnt, ll; FOR(i,2,pos) { if (xing[i]-xing[i-1]<=1) continue; sum=num=l=ll=cnt=0; FOR(j,xing[i-1]+1,xing[i]-1) { if (str[j]=='?') { ++num; if (l) node[++cnt].x=sum, node[cnt].len=l; node[cnt].flag=0, sum=l=0; } else { sum=sum*MOD+str[j]; ++l; if (num) node[++cnt].x=num, node[cnt].flag=1, num=0; } } if (l) node[++cnt].x=sum, node[cnt].len=l; node[cnt].flag=0, sum=l=0; if (num) node[++cnt].x=num, node[cnt].flag=1, num=0; while (bg<=ed&&!check(bg,cnt)) ++bg; if (!check(bg,cnt)) return false; FOR(i,1,cnt) if (node[i].flag) ll+=node[i].x; else ll+=node[i].len; bg+=ll; } return true; } int main () { hs[0]=1; FO(i,1,N) hs[i]=hs[i-1]*MOD; while (~scanf("%s%s",ss+1,str+2)) { pos=0; bg=1; ed=strlen(ss+1); for (int i=1; ss[i]; ++i) shs[i]=shs[i-1]*MOD+ss[i]; int l=1, r=strlen(str+1); str[1]='*'; str[++r]='*'; int flag=0; while (str[l]!='*'&&l<=r&&bg<=ed) { if (str[l]=='?'||str[l]==ss[bg]) ++l, ++bg; else {flag=1; break;} } while (str[r]!='*'&&l<=r&&bg<=ed) { if (str[r]=='?'||str[r]==ss[ed]) --r, --ed; else {flag=1; break;} } if (flag) {puts("NO"); continue;} FOR(i,l,r) if (str[i]=='*') xing[++pos]=i; if (pos==0&&bg<=ed) {puts("NO"); continue;} puts(sol()?"YES":"NO"); } return 0; }
D.Trigonometric Function(数学)
这题要放在高中我一定写的出来。。。
cos(na)化出来是一堆cos(a)和sin(a),sin(na)也是差不多。由余弦定理有cos(A)=(b^2+c^2-a^2)/(2*a*b).这tm显然是有理数。
那么只需要判断sinA,sinB,sinC是不是有理数即可。因为sinA=sqrt(1-cosA*cosA).化简这个式子然后就只需要判断一个东西是不是完全平方数就行了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-4 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; inline int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=20005; //Code begin... int main () { LL a, b, c, n, m, k; int T; scanf("%d",&T); while (T--) { scanf("%lld%lld%lld%lld%lld%lld",&a,&b,&c,&n,&m,&k); LL A=4*b*b*c*c-(b*b+c*c-a*a)*(b*b+c*c-a*a); LL B=4*a*a*c*c-(a*a+c*c-b*b)*(a*a+c*c-b*b); LL C=4*a*a*b*b-(a*a+b*b-c*c)*(a*a+b*b-c*c); LL a1=(LL)sqrt(A), b1=(LL)sqrt(B), c1=(LL)sqrt(C); if (a1*a1==A&&b1*b1==B&&c1*c1==C) puts("YES"); else puts("NO"); } return 0; }
F.Sleeping(DP)
将第0分钟设置为睡觉，第n+1分钟也设置为睡觉。那么题目就是求选出n个睡觉的时刻。使得它们两两间隔要么为0，要么大于l。
然后状态dp[i][j]表示前i分钟睡了j次且第i分钟在睡觉时可以浪费的最少技能点。
那么有dp[i][j]=max(dp[i-1][j-1],dp[i-k][j-1])+a[i],(k>=l+1);
最后统计一下睡觉次数>=m浪费的最少技能点就行了。复杂度O(n^2).
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-4 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; inline int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=1005; //Code begin... int a[N], dp[N][N]; int main () { int n, m, l; while (~scanf("%d%d%d",&n,&m,&l)) { int sum=0, ans=INF; FOR(i,0,n+2) FOR(j,0,n+2) dp[i][j]=INF; FOR(i,1,n) scanf("%d",a+i), sum+=a[i]; a[n+1]=0; dp[0][1]=0; FOR(j,2,n+2) { int mi=INF; FOR(i,j-1,n+1) { if (i-l-1>=0) mi=min(mi,dp[i-l-1][j-1]); dp[i][j]=min(mi+a[i],dp[i-1][j-1]+a[i]); } } FOR(i,m,n) ans=min(ans,dp[n+1][i+2]); printf("%d\n",sum-ans); } return 0; }
H.Replica Placement(树形DP)
求满足树上每个点要求的最小代价。可以用树形DP来求出。实际上一个点的副本只与它最近的祖先有关系。那么我们可以从根dfs并记录当前的距离。
如果距离可以满足点的限制，那么有两种选择，第一种这个点不作为副本，第二种这个点作为副本。但是这样dfs会超时。因为没有保存已经求出的东西。
令dp[x][y]表示x节点以y节点为副本的最小代价。那么之后就好搞了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi acos(-1.0) # define eps 1e-3 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=1205; //Code begin... struct Node{int q, s;}node[N]; struct Edge{int p, next, w;}edge[N]; int head[N], cnt=1, root, dp[N][N]; void add_edge(int u, int v, int w){ edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++; } void init(){mem(head,0); mem(dp,-1); cnt=1;} int dfs(int x, int w, int fa){ if (~dp[x][fa]) return dp[x][fa]; int res=0; for (int i=head[x]; i; i=edge[i].next) { int v=edge[i].p; if (node[v].q<w+edge[i].w) res+=(dfs(v,0,v)+node[v].s); else res+=min(dfs(v,w+edge[i].w,fa),dfs(v,0,v)+node[v].s); } return dp[x][fa]=res; } int main () { int T, n, fa, s, q, w; T=Scan(); while (T--) { init(); n=Scan(); FOR(i,1,n) { fa=Scan(); node[i].q=Scan(); node[i].s=Scan(); w=Scan(); if (fa) add_edge(fa,i,w); else root=i; } printf("%d\n",dfs(root,0,root)); } return 0; }
I.Triple(补集思想)
如果把数字看成点，两个数字如果互质则连一条红边，否则连一条蓝边。那么题目就是问同色三角形有多少个。
考虑补集。那么就是求非同色三角形的个数。
对于每个顶点，统计从这个顶点出发的两条边颜色都不一样的种数。然后对于所有顶点的种数/2就是非同色三角形的个数。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-4 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; inline int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=805; //Code begin... int a[N], num[N]; int gcd(int x, int y){return x==0?y:gcd(y%x,x);} int main () { int T, n; scanf("%d",&T); while (T--) { mem(num,0); scanf("%d",&n); LL ans=n*(n-1)*(n-2)/6, res=0; FOR(i,1,n) scanf("%d",a+i); FOR(i,1,n) FOR(j,i+1,n) if (gcd(a[i],a[j])!=1) ++num[i], ++num[j]; FOR(i,1,n) res+=num[i]*(n-1-num[i]); printf("%lld\n",ans-res/2); } return 0; }
J.Sudoku(Dancing Links)
学完之后就是模板题了。求n阶数独的解的存在性，判断下多解就行了。
# include <cstdio> # include <cstring> # include <cstdlib> # include <iostream> # include <vector> # include <queue> # include <stack> # include <map> # include <set> # include <cmath> # include <algorithm> using namespace std; # define lowbit(x) ((x)&(-x)) # define pi 3.1415926535 # define eps 1e-4 # define MOD 1000000007 # define INF 1000000000 # define mem(a,b) memset(a,b,sizeof(a)) # define FOR(i,a,n) for(int i=a; i<=n; ++i) # define FO(i,a,n) for(int i=a; i<n; ++i) # define bug puts("H"); # define lch p<<1,l,mid # define rch p<<1|1,mid+1,r # define mp make_pair # define pb push_back typedef pair<int,int> PII; typedef vector<int> VI; # pragma comment(linker, "/STACK:1024000000,1024000000") typedef long long LL; inline int Scan() { int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f; } void Out(int a) { if(a<0) {putchar('-'); a=-a;} if(a>=10) Out(a/10); putchar(a%10+'0'); } const int N=16; //Code begin... const int MaxN=N*N*N+10, MaxM=N*N*4+10, maxnode=MaxN*4+MaxM+10; int n, Pn; char g[20][20], G[20][20], gg[20][20]; int to_char(int x){return x<10?x+'0':x-10+'A';} struct DLX{ int n, m, size; int U[maxnode], D[maxnode], R[maxnode], L[maxnode], Row[maxnode], Col[maxnode]; int H[MaxN], S[MaxM], ansd, ans[MaxN]; void init(int _n, int _m){ n=_n; m=_m; FOR(i,0,m) S[i]=0, U[i]=D[i]=i, L[i]=i-1, R[i]=i+1; R[m]=0; L[0]=m; size=m; FOR(i,1,n) H[i]=-1; } void Link(int r, int c){ ++S[Col[++size]=c]; Row[size]=r; D[size]=D[c]; U[D[c]]=size; U[size]=c; D[c]=size; if (H[r]<0) H[r]=L[size]=R[size]=size; else R[size]=R[H[r]], L[R[H[r]]]=size, L[size]=H[r], R[H[r]]=size; } void remove(int c){ L[R[c]]=L[c]; R[L[c]]=R[c]; for (int i=D[c]; i!=c; i=D[i]) for (int j=R[i]; j!=i; j=R[j]) U[D[j]]=U[j], D[U[j]]=D[j], --S[Col[j]]; } void resume(int c){ for (int i=U[c]; i!=c; i=U[i]) for (int j=L[i]; j!=i; j=L[j]) ++S[Col[U[D[j]]=D[U[j]]=j]]; L[R[c]]=R[L[c]]=c; } int Dance(int d){ if (R[0]==0) { FO(i,0,d) G[(ans[i]-1)/Pn/Pn][(ans[i]-1)/Pn%Pn]=to_char((ans[i]-1)%Pn+1); return 1; } int c=R[0]; for (int i=R[0]; i!=0; i=R[i]) if (S[i]<S[c]) c=i; remove(c); int res=0; for (int i=D[c]; i!=c; i=D[i]) { ans[d]=Row[i]; for (int j=R[i]; j!=i; j=R[j]) remove(Col[j]); res+=Dance(d+1); if (res>=2) return 2; for (int j=L[i]; j!=i; j=L[j]) resume(Col[j]); } resume(c); return res; } }dlx; void place(int &r, int &c1, int &c2, int &c3, int &c4, int i, int j, int k){ r=(i*Pn+j)*Pn+k; c1=i*Pn+j+1; c2=Pn*Pn+i*Pn+k; c3=Pn*Pn*2+j*Pn+k; c4=Pn*Pn*3+((i/n)*n+(j/n))*Pn+k; } int main () { while (~scanf("%d",&n)) { Pn=n*n; FO(i,0,Pn) scanf("%s",g[i]); dlx.init(Pn*Pn*Pn,Pn*Pn*4); int r, c1, c2, c3, c4; FO(i,0,Pn) FO(j,0,Pn) FOR(k,1,Pn) if (g[i][j]=='.'||g[i][j]==to_char(k)) { place(r,c1,c2,c3,c4,i,j,k); dlx.Link(r,c1); dlx.Link(r,c2); dlx.Link(r,c3); dlx.Link(r,c4); } int ans=dlx.Dance(0); if (ans==0) {puts("No Solution"); continue;} else if (ans==2) {puts("Multiple Solutions"); continue;} int flag=1; FO(i,0,Pn) FO(j,0,Pn) gg[i][j]=G[i][j]; FO(i,0,Pn) { FO(j,0,Pn) { if (g[i][j]=='.') continue; char tmp=g[i][j]; g[i][j]='.'; dlx.init(Pn*Pn*Pn,Pn*Pn*4); FO(i,0,Pn) FO(j,0,Pn) FOR(k,1,Pn) if (g[i][j]=='.'||g[i][j]==to_char(k)) { place(r,c1,c2,c3,c4,i,j,k); dlx.Link(r,c1); dlx.Link(r,c2); dlx.Link(r,c3); dlx.Link(r,c4); } ans=dlx.Dance(0); g[i][j]=tmp; if (ans==1) {puts("Not Minimal"); flag=0; break;} } if (!flag) break; } if (flag) { FO(i,0,Pn) {FO(j,0,Pn) putchar(gg[i][j]); putchar('\n');} } } return 0; }
